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Let $X_1,...,X_n$ be a sample of i.i.d. random variables with mean 0 and finite variance, and let $\bar X$ denote the sample mean. How to show $E(|\bar X|)=O(1/\sqrt{n})$? Thanks.

Justin
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  • You need to assume a finite variance. In that case apply the central limit theorem. – herb steinberg Jul 06 '18 at 01:21
  • @ Justin This is trivial if variance is finite and highly non-trivial (probably false) otherwise. I will give a one line proof if you tell me that variance is finite. – Kavi Rama Murthy Jul 06 '18 at 05:41
  • @herbsteinberg I understand that with finite variance, $|\bar X|$ converges in distribution to a half-normal distribution, which has a mean of $O(1/\sqrt{n})$, but I am not very sure how to go from the convergence in distribution to the asymptotic bound. – Justin Jul 19 '18 at 20:42
  • @ Justin Wouldn't the proof look very much like the proof without absolute value? – herb steinberg Jul 19 '18 at 21:21
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    @herbsteinberg Without the absolute value $E(\bar X)=0$, so there is no need to look at the asymptotic bound. The central limit theorem is in the form of convergence in distribution. I am not sure how to go from the convergence in distribution to the asymptotic bound. – Justin Jul 20 '18 at 22:04

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This is actually Exercise 1.1.3 in Vershynin's "high-dimensional probability":

Let $X_1, X_2, \ldots$ be a sequence of iid random variables with mean $\mu$ and a finite variance. Then, $${\rm I\!E}\left|\frac{1}{N}\sum_{i=1}^{N}X_i - \mu\right| {}={} O\left(\frac{1}{\sqrt{N}}\right).$$

Without loss of generality we shall assume that $\mu=0$.

Firstly, we will get rid of that annoying absolute value as follows: Let us define $$\bar{X}_N = \frac{1}{N}\sum_{i=1}^{N}X_i.$$ Then, $${\rm I\!E}|\bar{X}_N| = {\rm I\!E} \sqrt{\bar{X}_N^2} \leq \sqrt{{\rm I\!E}[\bar{X}_N^2]},$$ where we used Jensen's inequality (using the fact that the square root is a concave function).

Now for ${\rm I\!E}[\bar{X}_N^2]$ we have $${\rm I\!E}[\bar{X}_N^2] = \frac{1}{N^2} {\rm I\!E}\left(\sum_{i=1}^{N}X_i\right)^2 = \frac{1}{N^2} {\rm I\!E}\left[\sum_{i=1}^{N}X_i^2\right] =\frac{1}{N^2} N\sigma^2 = \frac{\sigma^2}{N}.$$

As a result $${\rm I\!E}\left|\frac{1}{N}\sum_{i=1}^{N}X_i\right| {}\leq{} \sqrt{{\rm I\!E}[\bar{X}_N^2]} = \frac{\sigma}{\sqrt{N}},$$ which proves the assertion.

I tried to prove it starting from the Lindeberg-Lévy CLT and trying to use an argument based on uniform integrability, but I couldn't make it and I tend to believe that it's not possible. I would be very interested in a proof that leverages CLT if possible.

Pantelis Sopasakis
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